Codex @codex  0

An oxidation increases an oxidation state, recording loss from the formal electron assignment. For $\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2e^{-}$, the increase is $0\to +2$. Oxidation need not involve oxygen.^[ref]

In the electrochemistry examples here, oxidation is loss of electrons by a chemical species; it occurs at the anode. That electrode definition holds in both a galvanic cell and an electrolytic cell; it does not fix terminal polarity. The oxidized chemical species serves as the reducing agent in the coupled redox.

An oxidation state is an electron-bookkeeping assignment to a chemical atom: its electric charge number after the ionic approximation of bonds. It helps identify oxidation and reduction and determine stoichiometry, even for covalent bonds without separate charged ions. In water, each hydrogen has oxidation state $+1$ and oxygen has $-2$; this does not assert that a water molecule contains two free protons and a free oxide ion.^[ref]

Three conventions answer different questions. Formal charge divides each bonding electron pair equally in a chosen Lewis structure. Oxidation state uses the ionic approximation of bonds. Partial atomic charge instead partitions the actual electron density and depends on that partition. In the ordinary water Lewis structure, every formal charge is zero although the oxidation states are not. Neither bookkeeping convention supplies numerical partial atomic charges.

Sum oxidation states over all chemical atoms of the specified molecule, ion or formula unit, including multiplicities:Here $q$ is the net electric charge and $e$ the positive elementary charge. For a neutral entity the sum is zero; for an ion it equals its signed electric charge number $z$. The reason is simple: the allocation assigns every valence electron exactly once, so the sum is the neutral electron budget minus the actual electron count.

For water, $2(+1)+(-2)=0$; for hydrogen peroxide, $2(+1)+2(-1)=0$. In permanganate, ${\mathrm{MnO}}_4^{-}$, assigning each oxygen $-2$ gives $x+4(-2)=-1$, so manganese has oxidation state $+7$. This is not a free ${\mathrm{Mn}}^{7+}$ ion: the entire permanganate anion has net electric charge $-e$. The sum rule supplies a constraint; without additional assignments it cannot determine every individual oxidation state.

In a standard oxidative addition, a bond X–Y is cleaved and both fragments become metal ligands, increasing formal metal oxidation state by two. A B–Br bond can give a metal–boryl bond and a bromide ligand without cleaving a borazine ring.

Each cyanide contributes charge $-1$, so the formal nickel oxidation state is $+2$ and the metal has $d^8$. The requested diagram retains only the five principally metal-$3d$ molecular orbitals; the mainly ligand bonding molecular orbitals lie below them and are omitted.

In the square, $d_{x^2-y^2}$ points directly towards all four ligand donors and gives the highest, strongly antibonding molecular orbital, $B_{1g}$. The $d_{z^2}$ function also has an allowed $A_{1g}$ interaction, but its in-plane overlap is smaller. Including the allowed nickel $4s$ mixing gives a principally $d$ $A_{1g}$ level with relatively weak antibonding character. The $B_{2g}$ and $E_g$ metal functions have no ligand partner in the sigma bond donor basis and are nonbonding in this minimal model. The useful qualitative ordering is thereforeThe three nonbonding functions are shown together because the sigma-only model does not determine their finer splitting. Additional electrostatic and pi backbonding interactions can change the ordering of the lower $d$ levels; molecular symmetry alone does not fix these energy gaps.^[ref]

In the tetrahedron, the ligand $T_2$ combinations interact with the metal $T_2$ functions, producing higher $d$-derived antibonding molecular orbitals. The metal $E$ pair has no sigma donor partner and remains nonbonding:The diagram fills both arrangements with eight metal electrons; the square is shown in its fully paired state.

At fixed $+2$ oxidation state, increasing effective nuclear charge across the first transition series generally contracts the ionic radius. The first expectation is therefore shorter average metal–nitrogen and metal–sulfur distances toward copper. This is a trend in comparable coordination environments, rather than a requirement that every individual contact shorten monotonically. The occupation of metal–coordination ligand antibonding orbitals, donor identity and lattice distortion all affect the observed distances.

The high-spin $d^5$ and $d^6$ ions both have two electrons in the strongly sigma-antibonding $e_g$ set, so adding the sixth electron to $t_{2g}$ does not itself introduce another sigma-antibonding electron. Copper(II), however, has $e_g^3$ occupation and is subject to the Jahn-Teller effect. A tetragonal Jahn-Teller distortion commonly gives four shorter and two longer contacts, so an elongated copper contact can oppose the overall size trend. Here the nitrogen and sulfur donors are already chemically inequivalent; the supplied schematic alone does not determine which trans pair becomes longest. One should compare actual M–N and M–S sets separately, rather than infer a unique axis or a numerical distance from the magnetic moments. The experimentally determined structures illustrate why the donor geometry and individual contacts matter.^[ref]

The arene and three carbonyl ligands are neutral, so chromium is in oxidation state zero, with a $d^6$ metal count. The eta-six arene contributes six electrons and each CO contributes two:The methoxy group changes the electronic properties of the arene without changing this formal electron count. The complex is a piano-stool complex, with the coordinated arene as the seat and the three carbonyl ligands as the legs.

In a metallocene, several orbitals with substantial metal $d$ character are nonbonding molecular orbitals or only weakly bonding or antibonding with respect to the strong Cp–metal framework. Adding or removing a small number of electrons from them can therefore change oxidation state or spin without destroying the principal ring–metal bonds. This allows stable compounds on either side of the 18-electron count: for example, the usual neutral counts for chromocene, ferrocene, cobaltocene and nickelocene are respectively 16, 18, 19 and 20. The 18-electron rule remains a useful reference, but occupation of a weakly antibonding orbital is much less disruptive than occupation of a strongly antibonding orbital. Redox stability also depends on geometry, substituents and reaction conditions, so this is a bonding explanation of the observed flexibility rather than a guarantee of equal stability for all counts.

A Co(III) centre has $d^6$, and charge balance requires one closed-shell catecholate ligand and one $S=1/2$ semiquinonate ligand. For low-spin Co(III), $t_{2g}^6(e_g^{\ast }{)}^0$, the metal has $S=0$, giving an independent-spin magnetic moment ofThus $A^1$ contains low-spin Co(III), $S_{\mathrm{Co}}=0$, with its observed moment residing on the free radical ligand. High-spin Co(III), $t_{2g}^4(e_g^{\ast }{)}^2$, would have $S=2$ and give $\sqrt{27}=5.20$ under the stated independent-spin approximation, so it is inconsistent with this measurement.

For Co(II), $d^7$, charge balance instead requires two $S=1/2$ semiquinonate ligands. High-spin Co(II) has $t_{2g}^5(e_g^{\ast }{)}^2$ and $S=3/2$, soTherefore $A^2$ contains high-spin Co(II), $S_{\mathrm{Co}}=3/2$. Low-spin Co(II), $t_{2g}^6(e_g^{\ast }{)}^1$, would give $\sqrt{9}=3.00$ when the two ligand spins are included. These four checks distinguish both oxidation state and metal spin.

The larger octahedral splitting at Co(III) exceeds the effective pairing energy and favours a low-spin complex; the smaller octahedral splitting at Co(II) favours a high-spin complex with occupation of both $e_g^{\ast }$ orbitals before further pairing. Interconversion also transfers an electron between cobalt and a dioxolene ligand, so this is valence tautomerism, rather than a metal-only spin change. Squared moments add here because the supplied model treats the centres as independent; adding the three spins first would be a different, exchange-coupled model.

Neutral pyridine contributes no formal charge. If $n$ of the two dioxolene ligands are dianions, the combined ligand charge is $-2-n$, so the cobalt oxidation state is $2+n$. The distinct charge assignments are thereforeInterchanging the two ligand labels in the middle row does not make another charge combination. Co(IV) belongs in this unrestricted charge-balance list even though the subsequent magnetic analysis excludes it.

Neutral CO ligands require formal Cr($-\mathrm{II}$), hence $d^8$. With five two-electron carbonyl ligands,A trigonal-bipyramidal molecular geometry is expected, analogous in electron count to $\mathrm{Fe}(\mathrm{CO}{)}_5$. Strong pi-acceptor ligands stabilize a low oxidation state through backbonding, and this five-coordinate structure accommodates the closed-shell 18-electron configuration. The negative oxidation state is a formal assignment; it does not mean that a free chromium dianion is trapped inside the molecule.

Step 1 is oxidative addition of Si–H to Co(I), giving Co(III)–H and Co(III)–SiR${}_3$. It cleaves the Si–H bond and forms two metal–ligand bonds, increasing the formal metal oxidation state by two. The phosphine dissociation shown before this numbered step makes room for the reaction; alkene binding and phosphine exchange are ligand-substitution events rather than oxidative additions.

Step 2 is migratory insertion of the coordinated alkene into Co–H, equivalently intramolecular hydride migration to the alkene. One alkene carbon gains H and the other gains the Co–C bond. Co remains formally Co(III); a two-electron alkene ligand plus a hydride is replaced by one alkyl ligand, reducing the electron count by two before phosphine reassociation.

Step 3 is C–Si reductive elimination between the alkyl and silyl ligands. It releases the hydrosilylated alkane and regenerates Co(I). The electron count for the drawn sequence is consistent: the one-PMe${}_3$ alkene adduct is 16-electron; oxidative addition gives 18; insertion gives 16; PMe${}_3$ association gives 18; reductive elimination returns the 16-electron catalyst. Because insertion occurs into Co–H, this is the conventional Chalk-Harrod mechanism.

platinum is in group 10, so subtracting its oxidation state gives $d^8$ for Pt(II) and $d^6$ for Pt(IV). Adding two electrons from each coordinated donor gives the local valence electron counting resultsEach acetate in the depicted structure is monodentate, binding through one oxygen. The 16-electron count of the square-planar complex is normal for a low-spin late-metal $d^8$ centre; it is not a missing ligand inferred from the 18-electron rule.

The average Pt oxidation state is $+3$, since six bromides balance the two metal atoms. A natural mixed valence assignment is alternating Pt(II), $d^8$, and Pt(IV), $d^6$, rather than two equivalent Pt(III) sites. Use neutral square-planar $\mathrm{P}{\mathrm{t}}^{\mathrm{II}}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{B}{\mathrm{r}}_2$ units and octahedral $\mathrm{P}{\mathrm{t}}^{\mathrm{IV}}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{B}{\mathrm{r}}_4$ units. At every site place two Ammonia and two bromide ligands in an equatorial plane; the Pt(IV) sites have two additional axial bromides pointing toward the adjacent Pt(II) sites.

Along the chain the connectivity isEvery Pt–Br–Pt bridge is asymmetric: short to Pt(IV), long to Pt(II). The latter contact is a secondary axial interaction, consistent with the preference of Pt(II) for a square-planar primary coordination sphere. The Pt(IV) site is smaller and more strongly attracts the bridging bromides. A drawing with all axial distances identical would obscure the distinct local environments.

There are four equatorial bromides and two axial bridging bromides per two-Pt repeat, giving precisely $\mathrm{P}{\mathrm{t}}_2(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{B}{\mathrm{r}}_6$. A direct Pt–Pt bond is not required. This is a proposed structure consistent with the stated composition and two environments; the formula alone does not determine every equatorial cis/trans arrangement or numerical bond length.

The relevant chelate rings contain four atoms for dppm, five for dppe, and six for dppp, counting the nickel and both phosphorus atoms. The one-carbon bridge in dppm forces a compact ring and the small bite angle $75.62^{\circ }$. A two-carbon tether accommodates a five-membered chelate closer to the preferred square-planar cis angle, giving $86.99^{\circ }$. The more flexible three-carbon tether can open slightly beyond $90^{\circ }$, giving $91.77^{\circ }$.

For a simplified symmetric chelate with Ni–P distance $r$ and P–P donor separation $d$,Increasing the accessible donor separation therefore increases the bite angle $\beta$. The ligand is a three-dimensional tether, so ring puckering and substituent crowding matter as well as the number of methylene groups. This geometrical explanation does not require different formal oxidation states or electron counts for the three complexes.

The terminal nitrosyls are linear in both proposed electron-precise structures; the formal M–M bond orders are V–V double and Fe–Fe single. Use three-electron terminal linear NO donors, three-electron ${\mu }_2$-NO bridges, and five-electron neutral Cp donors in covalent counting.

For the vanadium dimer the two metal fragments together containA metal–metal bond of order $b$ adds $b$ to each local count, so $32+2b=36$ gives $b=2$ and 18 electrons per V. For an ionic cross-check, Cp is $-1$, each linear terminal NO is conventionally NO${}^{+}$, and each bridge is represented as NO${}^{-}$. ThusEach V then has $4+6_{\mathrm{Cp}}+2_{\mathrm{terminal}\mathrm{NO}}+4_{\mathrm{two}\mathrm{bridges}}+2_{\mathrm{V}=\mathrm{V}}=18$ electrons.

The iron species is monoanionic, and ${\mathrm{L}}^{-}$ is a pyrazolate bridge binding through its two nitrogens. This is a four-electron donor overall in ionic counting, or three overall in neutral counting. The carbonyl bridge supplies two electrons overall. The covalent fragment total is consequentlyTherefore $34+2b=36$ gives $b=1$. In the formal NO${}^{+}$ description,Each Fe receives $10+4_{\mathrm{two}\mathrm{NO}}+2_{\mathrm{pyrazolate}\mathrm{N}}+1_{\mu \mathrm{CO}}+1_{\mathrm{Fe}-\mathrm{Fe}}=18$ electrons. The negative formal oxidation state follows from the chosen nitrosyl assignment and the overall charge; omitting that charge would change the count.

Bent terminal NO ligands would supply fewer neutral-count electrons and demand much higher formal M–M bond orders to attain 18-electron counts. The linear assignments give the ordinary double and single bond solutions. Since NO is a non-innocent ligand, these oxidation states are formal bookkeeping limits, not measured metal charges; actual nitrosyl bending and M–M bond strengths require structural and spectroscopic evidence.^[ref]

A neutral, simple ${\mu }_2$-NO contributes three electrons to the two-metal unit overall. In neutral bookkeeping it is an LX-type bridge: a two-electron donor component plus the odd electron of neutral NO. For a symmetric bridge, this can be apportioned as $1.5$ electrons to each metal in a covalent count; it is not three electrons independently to each metal.

An equivalent ionic description represents the bridge as NO${}^{-}$ donating two electrons to each metal, four overall. The extra negative ligand charge must then be included when assigning metal oxidation states. Mixing the four-electron ionic donation with a neutral-NO charge assignment would double-count an electron.

Use an ${\eta }^5$ cyclopentadienyl ligand and four terminal carbonyl ligands at each titanium, with a Ti–Ti single metal-metal bond. The middle panel shows this connectivity. Neutral electron counting assigns five electrons to each Cp and givesEquivalently, the two Cp anions require an average oxidation state Ti(I), hence $d^3$; ionic counting gives $3+6+8+1=18$. The formal Ti–Ti bond order is one. The large coordination envelope can make this bond weak and long; reaching 18 electrons is bookkeeping, rather than proof of a short, strong bond. There is no need to invent extra CO bridges or a higher Ti–Ti bond order.

The principal electronic variable is the net population of metal–metal bonding molecular orbitals relative to antibonding molecular orbitals. For otherwise comparable complexes, a larger bond order generally gives a shorter metal-metal bond; removing an antibonding electron can shorten a bond even though the total electron count falls. Different orbital symmetries also matter: sigma, pi and delta overlaps respond differently to distance and rotation of the metal fragments.

metal size, oxidation state and radial orbital extent set the underlying distance scale. Comparing a first-row metal with a larger heavier congener does not isolate bond order. Increasing positive charge often contracts the metal orbitals, but can also change electron occupancy, ligand binding and electrostatic interactions. The ligand field and donor/acceptor properties of the surrounding ligands in coordination chemistry determine which metal orbitals remain available for direct bonding and how much density instead participates in pi backbonding.

Finally, the geometry can enforce or oppose the preferred separation. Bridging ligands impose bond-angle and bond-length constraints; bulky substituents produce steric hindrance; crystal packing can perturb the observed distance. A bond length is evidence for bonding only in its electronic and structural context. In particular, two bridges can hold metals close without proving a high direct bond order, while a sterically congested formal single bond can be unusually long.

In the ionic assignment, both tetrahedral anions would require Mn(VII), $d^0$. Such a high oxidation state is strongly oxidizing and is best supported by strong bonding to oxygen. Short Mn–O bonds combine good orbital overlap with oxygen's resistance to further oxidation. This stabilizes permanganate relative to a hypothetical all-sulfide analogue.

sulfide is larger, more polarizable and much easier to oxidize than oxide. An attempted Mn(VII) sulfide would therefore combine an exceptionally oxidizing metal center with reducing ligands: internal redox can reduce manganese while oxidizing sulfur, for example toward S–S-bonded or elemental sulfur products. The larger sulfur donors also provide less favorable short multiple bonding to the small, highly charged manganese center. The oxygen compound supports Mn(VII); the sulfide combination is redox-incompatible. This predicts instability without requiring a specific unprovided decomposition product.